Question

If ${}^{28}{C}_{2r}{:}^{24}{C}_{2r-4}=225:11$,find 2r.

Answer 1

${}^{28}{C}_{2r}{:}^{24}{C}_{2r-4}=225:11$

Let $2r=n$ ${}^{28}{C}_n{:}^{24}{C}_{n-4}=225:11$

$\frac{28\times 27\times \mathrm{26.....}(28-n+1)}{\left(n\right)(n-1).....2.1}=\frac{(n-4)1}{24\times 23\times 22=(24-n+4r)}$

$\frac{28\times 27\times 26\times 25}{n(n-1)(n-2)(n-3)}=1=225=11$

$\frac{28\times 3\times 26}{n(n-1)(n-2)(n-3)}=\frac{1}{11}$

$n(n-1)(n-2)(n-3)=28\times 3\times 26\times 11=11\times 12\times 13\times 14$

$n=11$

$n=2r$

$2r=11$

$r=55$